# Fourier transform of the G-lesser function.

+ 1 like - 0 dislike
26 views

I am studying Kadanoff & Baym's book Quantum Statistical Mechanics and I am stuck a one point.

The are considering a system of non-interacting particles, (let's say fermions to not having to write both signs), and are then considering the G-lesser function:

$$G^{<} (1, 1') = i \left< \psi^\dagger (1') \psi(1) \right> ,$$

where $1 = \mathbf{r}_1, t_1$ and similarly for $1'$.

Since the Hamiltonian has rotational and translational symmetry they argue that the Green's function above only depends on $| \mathbf{r}_1 - \mathbf{r}_{1'} |$. Also, since the Hamiltonian is time independent the Green's function should only depend on the time difference $t_1 - t_1'$. All this seems fine and I think I have sucessfully convinced myself of these facts by considering e.g. the translation operator.

However, they then define the Fourier transform as

$$G^{<} ( \mathbf{p}, \omega) = - i \int d^3 r \int dt e^{-i \mathbf{p} \cdot \mathbf{r} + i \omega t} G^{<}(\mathbf{r}, t),$$

where we now use $\mathbf r = \mathbf r_1 - \mathbf r_2$ and similarly for $t$. Now come the claim that I cannot really see. They say that, due to the invariances I talked about above, we have

$$G^{<}(\mathbf{p} , \omega) = \int dt \frac{e^{i\omega t}}{V} \left< \psi^\dagger(\mathbf{p}, 0) \psi(\mathbf{p}, t) \right>$$

where $V$ is the volume of the system. Can someone please explain how this follows from the above? If I naively try to calculate this I instead get

$$G^{<} ( \mathbf{p}, \omega) = - i \int d^3 r \int dt e^{-i \mathbf{p} \cdot \mathbf{r} + i \omega t} G^{<}(\mathbf{r}, t) \\ = - i \int d^3 r \int dt e^{-i \mathbf{p} \cdot \mathbf{r} + i \omega t} \left< \psi^\dagger(\mathbf{0}, 0) \psi(\mathbf{r}, t ) \right>,$$

which would only give the Fourier transform of the annihilation operator.

I have tried to redo everything carefully from scratch and I have found for example the relation

$$G^{<} ( \mathbf{p}, \omega, \mathbf{p}', \omega' ) = 2\pi V \delta(\omega - \omega') \delta_{\mathbf{p}, \mathbf{p}'} G^{<} ( \mathbf{p}, \omega).$$

However, I cannot really get the factor $1/V$ which was obtained in the book.

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysics$\varnothing$verflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.